Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 5 · Chapter companion part
Combine unlike algebraic fractions with justified common denominators and handle whole-numerator subtraction.
Class 8 English Version: part B of the chapter 5 companion. Eight sections, 30 written tasks and a notebook project. Selected companion coverage, not complete textbook coverage; use all four parts. Numerical/coefficient answers are checked; equivalent symbolic forms, domains and proofs need guardian or teacher review.
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For a shared nonzero D, A/D + B/D = (A+B)/D. Different denominators need equivalent common-denominator forms first. The original strip model shows 1/2 + 1/3: repartition into sixths to obtain 3/6 + 2/6 = 5/6. Adding denominators would give 2/5, a different value. The same rule applies to algebraic denominators.
For 1/(2x) + 1/(3x), use 6x, requiring x≠0. Numerators become 3 and 2, so the sum is 5/(6x). For x/6 + x/4, use 12 and obtain 2x/12 + 3x/12 = 5x/12. A common denominator changes the form, not the value; do not multiply a numerator without making the same denominator change.
For 1/(x−2) + 1/(x+2), D=(x−2)(x+2)=x²−4 and the numerator is 2x. A cubic difference factors as x³−y³=(x−y)(x²+xy+y²). Thus the denominators x−y, x²+xy+y² and x³−y³ can all use x³−y³; the first two are not multiplied in again after the cubic already contains them.
For 1/[(x−1)(x−2)] + 1/[(x−2)(x−3)] + 1/[(x−3)(x−4)], use D=(x−1)(x−2)(x−3)(x−4). The new numerators are (x−3)(x−4), (x−1)(x−4) and (x−1)(x−2). Add them to 3x²−15x+18 = 3(x−2)(x−3). The final result is 3/[(x−1)(x−4)], but x=1,2,3,4 all stay excluded.
A/D − B/D = (A−B)/D. For (2x+1)/(x−1) − (x−3)/(x−1), subtract (x−3): 2x+1−x+3 = x+4. The result is (x+4)/(x−1), x≠1. Forgetting brackets incorrectly gives x−2. After an unlike-denominator rewrite the second numerator may be a product; subtract that entire product too.
For x/(x−y)² − (x+y)/(x²−y²), the second fraction reduces to 1/(x−y), on x≠y,−y. Rewriting gives [x−(x−y)]/(x−y)² = y/(x−y)². Combining first over (x−y)²(x+y) gives the same result. Preserve the original exclusions even when the result has a shorter denominator.
For (x−y)/[(y+z)(z+x)] + (y−z)/[(z+x)(x+y)] + (z−x)/[(x+y)(y+z)], use D=(x+y)(y+z)(z+x). The numerators become x²−y², y²−z² and z²−x². Their sum is zero. The expression equals zero only where D≠0; a zero result does not authorise division by zero in the original expression.
Source example 5(c) on printed page 82 asks 1/(a−2)+(a−2)/(a²+2a+4). The solution on page 83 instead uses a+2 in the second numerator. As stated, the common numerator is 2a²−2a+8 over a³−8. For the changed version it is 2a(a+1) over a³−8. At a=4 these give 4/7 and 5/7 respectively. They cannot be blended; label which expression you are solving.
To subtract (x−3)/(x−1) from (2x+1)/(x−1), keep the common denominator and write [(2x+1)−(x−3)]/(x−1). Collect to (x+4)/(x−1), x≠1. At x=2 the original is 5−(−1)=6, agreeing with the result.
Adding denominators is not the fraction-addition rule. A common denominator lets numerators count pieces of one size.
Example 5(c) changes a−2 to a+2 between printed pages 82 and 83. Both worked versions are kept distinct in this companion.
The numerator operation changes, while the common denominator stays.
| Expression | Common numerator | Result/domain |
|---|---|---|
| 1/(x−2)+1/(x+2) | (x+2)+(x−2)=2x | 2x/(x²−4), x≠±2 |
| 1/(x−2)−1/(x+2) | (x+2)−(x−2)=4 | 4/(x²−4), x≠±2 |
| (2x+1)/(x−1)−(x−3)/(x−1) | 2x+1−(x−3)=x+4 | (x+4)/(x−1), x≠1 |
The subtraction bracket changes both signs of x−3. At x=2 the last expression is 6, not the value obtained by dropping that bracket.
Original equal-length strips show one half and one third of the same whole. Repartitioning changes the count of parts, not their total size.
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Cancel only after adding all complete new numerators.
1/[(x−1)(x−2)] + 1/[(x−2)(x−3)] + 1/[(x−3)(x−4)] = [3x²−15x+18]/[(x−1)(x−2)(x−3)(x−4)] = 3/[(x−1)(x−4)]
Do not multiply the three original denominators blindly: that repeats factors unnecessarily.
Source example 5(c), printed pages 82–83, switches a−2 to a+2. These are separate calculations.
Stated: 1/(a−2)+(a−2)/(a²+2a+4) = (2a²−2a+8)/(a³−8) Changed: 1/(a−2)+(a+2)/(a²+2a+4) = 2a(a+1)/(a³−8)
The companion preserves the source discrepancy and explains both. Do not quietly repair the question while using the other answer.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original numerical bridge to the algebraic common-denominator method; cells represent exact equal parts of one whole, not measured data.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original symbolic relationship guide; arrows are reasoning links, not a geometric scale.
Show your working on paper. Check your answer, then choose Next.
5
Repartition: 3/6 + 2/6 = 5/6.
5
The common numerators are 3 and 2, so n = 5.
5
2x + 3x = 5x over 12.
2
The numerator is (x+2)+(x−2)=2x.
1
1/2 + 2/4 = 1.
2
The denominators exclude 2 and −2.
4
1 − (−3) = 4.
6
5 − (−1) = 6, also (2+4)/(2−1).
4
(x+2)−(x−2)=4.
0
The first difference is exactly 4/(x²−4), so the total is zero.
0.25
The reduced result is y/(x−y)² = 1/4.
0
All pair sums are nonzero and the common numerator telescopes to zero.
3
Each missing-factor product contributes x²; total 3x²−15x+18.
32
2a²−2a+8 = 32, over a³−8 = 56.
40
2a(a+1) = 40, over 56. It is a different expression.
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Divide one whole into six equal parts. Half is three sixths, a third is two sixths and their sum is five sixths. The denominator names the common part size.
Draw equal wholes, labelled sixths and a total of five shaded cells.
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(4y²+5x)/(6x²y³), for x,y≠0. Multiply the first numerator by 2y² and the second by x.
Check all missing powers and avoid combining unlike numerator terms.
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2(x²+4)/(x³−8), for x³−8≠0. The common numerator is x²+2x+4+x(x−2)+4=2x²+8.
Use the cubic factorisation and show cancellation of the linear terms.
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2a(3a+5)/[(a+4)(a²−1)], for a≠−4,−1,1. The common numerator is 3a(a+1)+2a(a+4)+a(a−1)=6a²+10a.
Factor every denominator before finding multipliers.
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(a²+b²+c²−ab−bc−ca)/(abc), for abc≠0. The new numerators are a(a−b), b(b−c), c(c−a).
Do not assume every cyclic sum is zero; read its actual denominator pattern.
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3/[(x−1)(x−4)], for x≠1,2,3,4. The common numerator 3x²−15x+18 factors to 3(x−2)(x−3).
Retain the two cancelled exclusions as well as the final denominator roots.
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(x+4)/(x−1), x≠1. The bracket −(x−3) contributes −x+3.
Require the subtraction bracket before collecting.
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8ab/(a²−16b²), with a−4b and a+4b nonzero. Subtract (a−4b)² from a²+16b² in the common numerator.
Check the middle square term and both exclusions.
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2y/(x²−y²), on the original domain x−y≠0 and x³+y³≠0. The second fraction reduces to 1/(x+y); retain its original restrictions.
Use sum-of-cubes factors and subtraction parentheses.
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32/(x⁴−16), for x≠2,−2. First difference is 4/(x²−4); subtract over (x²−4)(x²+4), whose numerator is 32.
Do not confuse the final denominator x²+4 with x²−4.
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Over D=(x+y)(y+z)(z+x), numerators become x²−y²,y²−z²,z²−x². Their sum is zero when D≠0.
Show each missing factor and cancellation; include the domain.
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Reducing the second fraction first gives 1/(x−y), then y/(x−y)². Combining over (x−y)²(x+y) gives y(x+y) in the numerator and the same result. Require x−y≠0 and x+y≠0.
The two routes must keep the same original domain.
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(2a²−2a+8)/(a³−8), for a≠2 over the reals. The second numerator contributes (a−2)². At a=4 the value is 4/7.
Identify the a−2 numerator and do not borrow the changed solution.
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2a(a+1)/(a³−8), for a≠2 over the reals. The second numerator contributes (a−2)(a+2)=a²−4. At a=4 the value is 5/7.
Label this as the changed expression, not the original question.
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The correct rule is A/D−B/D, with D≠0. For A=8,B=2,D=3, both correct forms give 2, while 8/3−2 gives 2/3.
Explain why D divides the whole numerator, including B.
Write an addition/subtraction audit using a strip model, a quadratic-denominator sum and a cyclic cancellation proof.
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This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 5, printed pages 77–100. It does not replace the complete chapter or its practical teaching.
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